Line Integrals of Scalar Fields - Calculus 3

1. Consider two curves: $C_1$ is the straight line from $(0,0,0)$ to $(1,1,1)$, and $C_2$ is the curve parameterized by $\mathbf{r}(t) = (t, t^2, t^3)$ for $0 \leq t \leq 1$. Both curves connect the same endpoints. If $f(x,y,z) = x^2 + y^2 + z^2$, which statement about the line integrals $\int_{C_1} f \, ds$ and $\int_{C_2} f \, ds$ is correct?
   Answer: The relationship depends on the specific values computed for each integral Line integrals of scalar fields are path-dependent and depend on both the function values along the path and the arc length element. For $C_1$: $\mathbf{r}_1(t) = (t,t,t)$ for $t \in [0,1]$, so $|\mathbf{r}_1'(t)| = \sqrt{3}$ and $f(\mathbf{r}_1(t)) = 3t^2$. Thus $\int_{C_1} f \, ds = \sqrt{3}\int_0^1 3t^2 \, dt = \sqrt{3}$. For $C_2$: $\mathbf{r}_2(t) = (t,t^2,t^3)$ for $t \in [0,1]$, so $|\mathbf{r}_2'(t)| = \sqrt{1 + 4t^2 + 9t^4}$ and $f(\mathbf{r}_2(t)) = t^2 + t^4 + t^6$. The integral $\int_{C_2} f \, ds = \int_0^1 (t^2 + t^4 + t^6)\sqrt{1 + 4t^2 + 9t^4} \, dt$ is more complex but computable. Option A is wrong because shorter arc length doesn't guarantee larger integral value. Option B is wrong because longer arc length doesn't guarantee larger integral value either - the function values matter. Option C is wrong because line integrals of scalar fields are path-dependent (unlike line integrals of conservative vector fields). Option D is correct because we need to compute both integrals to determine their relationship.
2. The work done by a force field $\mathbf{F}(x,y,z) = (y, x, z^2)$ in moving a particle along the curve $\mathbf{r}(t) = (\cos t, \sin t, t)$ from $t = 0$ to $t = \pi$ is equal to the line integral of which scalar field over the same curve?
   Answer: $f(x,y,z) = (\sin t)(-\sin t) + (\cos t)(\cos t) + t^2(1)$ Work is computed as $\int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt$. Here, $\mathbf{F}(\mathbf{r}(t)) = (\sin t, \cos t, t^2)$ and $\mathbf{r}'(t) = (-\sin t, \cos t, 1)$. The dot product is $\mathbf{F} \cdot \mathbf{r}' = (\sin t)(-\sin t) + (\cos t)(\cos t) + t^2(1) = -\sin^2 t + \cos^2 t + t^2 = \cos(2t) + t^2$. This equals the line integral $\int_C f \, ds$ where $f$ is the scalar field that, when evaluated along the curve and multiplied by $ds/dt = |\mathbf{r}'(t)|$, gives the same integrand. But more directly, the work integral $\int_0^\pi (\sin t)(-\sin t) + (\cos t)(\cos t) + t^2(1) \, dt$ matches option D exactly. The other options incorrectly represent the relationship between work and line integrals of scalar fields.

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